【Codeforces238E】Meeting Her

题意:

  • 有一张 n 个点,m条边的有向图,有 k 种巴士,每种巴士有一个起点si和终点 ti ,每一个时刻,巴士都会等概率选择一条两点间的最短路径。
  • 询问一次从 a b最坏情况下最少需要乘坐几种巴士。
  • n100,mn×(n1),k100

题解:

  • 首先我们需要知道每个巴士的路径上的必经点。
  • 然后我们可以通过用 dp[i] 表示从 i b最坏情况的最少次数。
  • 不断地尝试用 k <script type="math/tex" id="MathJax-Element-12">k</script>种巴士来更新答案,到不能更新时停止。
  • 每次更新时采用记忆化搜索。
  • 保证上车点是必经点的情况下,尝试更新。

代码:

#include <bits/stdc++.h>
#define gc getchar()
#define ll long long
#define N 109
using namespace std;
int f[N][N],n,m,s,t,b_num,dp[N],vis[N],Dp[N],T;
map<int,int> mp;
struct bus
{
    int s,t;
    int vis[N];
    vector<int> q;
    void st()
    {
        memset(vis,0,sizeof(vis));
        int cnt=0;
        for (int now:q) vis[now]=f[s][now]+1;
    }
}b[N];
int dfs(int x,int pos)
{
    if (vis[x]==T) return Dp[x];
    vis[x]=T;
    int ret=-1;
    for (int i=1;i<=n;i++)
        if (f[x][i]==1&&1+f[i][b[pos].t]==f[x][b[pos].t]) ret=max(ret,dfs(i,pos));
    if (ret==-1) ret=1e9;
    ret=min(ret,dp[x]);
    return Dp[x]=ret;
}
int main()
{
    ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    cin>>n>>m>>s>>t;
    memset(f,0x3f,sizeof(f));
    for (int i=1;i<=n;i++) f[i][i]=0;
    for (int i=1;i<=m;i++)
    {
        int x,y;
        cin>>x>>y;
        f[x][y]=1;
    }
    cin>>b_num;
    for (int i=1;i<=b_num;i++)
        cin>>b[i].s>>b[i].t;
    for (int k=1;k<=n;k++)
        for (int i=1;i<=n;i++)
            for (int j=1;j<=n;j++)
                if (f[i][k]+f[k][j]<f[i][j]) f[i][j]=f[i][k]+f[k][j];
    for (int i=1;i<=b_num;i++)
    {
        mp.clear();
        if (f[b[i].s][b[i].t]>n) continue;
        for (int j=1;j<=n;j++)
            if (f[b[i].s][b[i].t]==f[b[i].s][j]+f[j][b[i].t]) mp[f[b[i].s][j]]++;
        for (int j=1;j<=n;j++)
            if (f[b[i].s][b[i].t]==f[b[i].s][j]+f[j][b[i].t]&&mp[f[b[i].s][j]]==1) b[i].q.push_back(j);
        b[i].st();
    }
    memset(dp,0x3f,sizeof(dp));
    dp[t]=0;
    int flag=1;
    while (flag)
    {
        flag=0;
        for (int i=1;i<=b_num;i++)
            for (int j=1;j<=n;j++)
                if (b[i].vis[j])
                {
                    T++;
                    int tmp=dfs(j,i)+1;
                    if (tmp<dp[j])
                    {
                        flag=1;
                        dp[j]=tmp;
                    }
                }
    }
    if (dp[s]>n) puts("-1");
    else cout<<dp[s]<<endl;
    return 0;
}
### Codeforces 887E Problem Solution and Discussion The problem **887E - The Great Game** on Codeforces involves a strategic game between two players who take turns to perform operations under specific rules. To tackle this challenge effectively, understanding both dynamic programming (DP) techniques and bitwise manipulation is crucial. #### Dynamic Programming Approach One effective method to approach this problem utilizes DP with memoization. By defining `dp[i][j]` as the optimal result when starting from state `(i,j)` where `i` represents current position and `j` indicates some status flag related to previous moves: ```cpp #include <bits/stdc++.h> using namespace std; const int MAXN = ...; // Define based on constraints int dp[MAXN][2]; // Function to calculate minimum steps using top-down DP int minSteps(int pos, bool prevMoveType) { if (pos >= N) return 0; if (dp[pos][prevMoveType] != -1) return dp[pos][prevMoveType]; int res = INT_MAX; // Try all possible next positions and update 'res' for (...) { /* Logic here */ } dp[pos][prevMoveType] = res; return res; } ``` This code snippet outlines how one might structure a solution involving recursive calls combined with caching results through an array named `dp`. #### Bitwise Operations Insight Another critical aspect lies within efficiently handling large integers via bitwise operators instead of arithmetic ones whenever applicable. This optimization can significantly reduce computation time especially given tight limits often found in competitive coding challenges like those hosted by platforms such as Codeforces[^1]. For detailed discussions about similar problems or more insights into solving strategies specifically tailored towards contest preparation, visiting forums dedicated to algorithmic contests would be beneficial. Websites associated directly with Codeforces offer rich resources including editorials written after each round which provide comprehensive explanations alongside alternative approaches taken by successful contestants during live events. --related questions-- 1. What are common pitfalls encountered while implementing dynamic programming solutions? 2. How does bit manipulation improve performance in algorithms dealing with integer values? 3. Can you recommend any online communities focused on discussing competitive programming tactics? 4. Are there particular patterns that frequently appear across different levels of difficulty within Codeforces contests?
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